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‘iexploit is awesome by shellcode’
‘lhasorlw lv dzhvrph eb vkhoofrgh’
‘lhasorlw lv dzhvrph eb vkhoofrgh’
‘lhasorlwlvdzhvrphebvkhoofrgh’
‘lhas orlw lvdz hvrp hebv khoo frgh’
public String encrypt(String plainText, int shift) {
StringBuilder code = new StringBuilder();
for (char c : plainText.toCharArray()) {
if (Character.isWhitespace(c)) {
code.append(c); //leave whitespaces the way they are
} else {
int s = ((c % 'A' + shift) % 26); //the encryption formula
if (s > 25) {
s = s - 26; //result is not in range 0...25
} else if (s < 0) {
s = s + 26; //result is not in range 0...25
}
code.append((char) (s + 'A')); //append resulting letter to the encrypted string
}
}
return code.toString();
}
public String decrypt(String code, int shift) {
return encrypt(code, -shift);
}
What I did for cracking a shift cipher was just determining the most frequent letter which should translate to 'e'. If it was wrong you can try the second most frequent letter for 'e' and so on.
This is a kind of frequency analysis too, but very simple and you don't need to read all 26 possibilities.
Good paper, Sh3llc0d3.
but also have less than 100% chance of one of them being the correct shift.
would not only increase coding time
but also have less than 100% chance of one of them being the correct shift.
No it hasn't a less chance. My way is the same as yours. I just give the messages a better order, so that the first is the message that is most probably correct.
For your bruteforcing the user has to read through 25 messages to determine the correct one. I give it a bit more convenience.would not only increase coding time
It is not complicate to count frequencies of single letters. This is done in ten minutes coding time.
where do they use this cipher? I mean advance ciphers are available so i don't think they use it anywhere,but really a VERY GOOD explanation
where do they use this cipher? I mean advance ciphers are available so i don't think they use it anywhere,but really a VERY GOOD explanation
public String encrypt(String plainText, int shift) {
StringBuilder code = new StringBuilder();
for (char c : plainText.toCharArray()) {
if (Character.isWhitespace(c)) {
code.append(c); //leave whitespaces the way they are
} else {
int s = ((c % 'A' + shift) % 26); //the encryption formula
if (s > 25) {
s = s - 26; //result is not in range 0...25
} else if (s < 0) {
s = s + 26; //result is not in range 0...25
}
code.append((char) (s + 'A')); //append resulting letter to the encrypted string
}
}
return code.toString();
}
public String decrypt(String code, int shift) {
return encrypt(code, -shift);
}